Question #1556c

1 Answer
Jun 23, 2016

#ΔH = "-208 kJ"#

Explanation:

The reaction is

#"CH"_4 + "2Cl"_2 → "CH"_2"Cl"_2 + "2HCl"#

Let's count the number of bonds in each substance.

#color(white)(l)"CH"_4color(white)(mll) = "4 C-H bonds"#
#color(white)(l)"2Cl"_2 color(white)(mll)= "2 Cl-Cl bonds"#
#color(white)(l)"CH"_2"Cl"_2 = "2 C-H bonds + 2 C-Cl bonds"#
#color(white)(l)"2HCl" color(white)(ml)= "2 H-Cl bonds"#

So, we have

#color(white)(mmmll)"CH"_4 + color(white)(m)"2Cl"_2color(white)(ll) → color(white)(mmll)"CH"_2"Cl"_2color(white)(mm) + color(white)(l)"2HCl"#
#color(white)(mmml)"4C-H" + color(white)(l)"2Cl-Cl" → color(white)(ll)"(2C-H + 2C-Cl)" + color(white)(l)"2H-Cl"#
#D"/kJ:"color(white)(l) "4×413"color(white)(ml) "2×242"color(white)(mml) "(2×413 + 2×328)" color(white)(ml)"2×431"#
#D"/kJ:"color(white)(ll) 1652 +color(white)(m) 484 color(white)(mmmml)"(826 + 656)"color(white)(ml) + color(white)(m)"862"#
#D"/kJ:" color(white)(mmm)2136color(white)(mmmmmmmmmmm) 2344#

#ΔH = ΣD_"reactants" -ΣD_"products" = "2136 kJ -2344 kJ" = "-208 kJ"#